Mathematics - 2017-01-21 (page 2 of 3)

Balarka Sen

15:00

You can definitely estimate a size. The point is by mean value theorem $\phi$ is Lipschitz: $|\phi(x) - \phi(y)| \leq C |x - y|$ where $C = \max_{x \in I} |\phi'(x)|$, $I$ being the interval where $|\phi'(-)| < 1$.

So let's see what Banach fixed point theorem says explicitly.

Secret

[Day 3: Trial number 12] Evaluate the integral such that the evaluation highlights as many of its symmetries as possible
$$\int x^n \sin x e^x dx$$
I hope I can stop making careless mistakes!

Balarka Sen

@Alessandro Wait, no, I didn't read what you said correctly. I don't understand the question: if $I$ is the nbhd of $\alpha$ in which $|\phi'| < 1$, it's precisely where the convergence is guaranteed...?

Why's that not a "relation"

Riccardo

I've a quick question about Steenrod Squares. I want to compute explicit generators of H_5(K(Z_2,2);Z_2), and according to UCT, H_5(K(Z_2,2);Z_2)=Z_2+Z_2. According to Serre's Theorem about the cohomology rings of K(Z_2,n), it should be generated by image of the fundamental class i in H^2(K(Z_2,2);Z_2) via iterated Steenrod Squares. In particular, I should be able to find 2 non zero composition of iterated Steenrod Squares evaluated in i. I'm only able to find one of them, Sq^2Sq^1(i)

Balarka Sen

I see, the theorem you quoted does not mention that. Strange.

Alessandro Codenotti

@Balarka I'm not sure, it's phrased in an horrible way in my textbook, I know that there is a nbhd $I_1$ of $\alpha$ in which convergence is guaranteed and that $|\phi'(\alpha)|<1$ (so by continuity there is a nbhd $I_2$ of $\alpha$ in which $|\phi'(\alpha)|<1$), what I'm trying to understand is whether $I_1=I_2$

I'm quite certain that $I_2\subseteq I_1$

Riccardo

15:06

the other sequences are zero evaluated on i, (1,1,1) or (1,2) or (3). Can't find another one sadly

Balarka Sen

$I_2$ can be any closed neighborhood of $\alpha$ contained in $I_1$.

Alessandro Codenotti

Hmm, that's weird. The theorem before that is basically a specific instance of the contraction theorem: if $\phi\in C^0([a,b])$, $\phi([a,b])\subseteq [a,b]$ and $\exists L<1$ such that $|\phi(x_1)-\phi(x_2)|\le L|x_1-x_2|$ then there is a unique fixed point $\alpha$ of $\phi$ in $[a,b]$ and the sequence $x_{k+1}=\phi(x_{k})$ converges to $\alpha$ for all $x_0\in[a,b]$

Balarka Sen

"Contained" was not strict. If $I_1$ is closed then $I_2 = I_1$ is totally fine.

Alessandro Codenotti

yeah so in general the nbhd with guaranteed convergence is at least as big as the one in which the derivative is bounded by $1$

Balarka Sen

right

Mary Star

15:17

Why is $\mathbb{Q}(\sqrt{2})\cup \mathbb{Q}(\sqrt{3})$ not a field?

Alessandro Codenotti

it doesn't contain $\sqrt{2}+\sqrt{3}$

Mary Star

@AlessandroCodenotti Ah ok. Thanks!!

Pichi Wuana

15:31

Can any right triangle be inscribed in a circle? I'm thinking the fact that a inscribed angle = 90° is pointing to the diameter.

DHMO

@PichiWuana yes, and you've just explained why.

Balarka Sen

ANY triangle can be inscribed in a circle

Pichi Wuana

Oh hahaha

DHMO

@BalarkaSen I was just about to say that

Alessandro Codenotti

for a right triangle a semicircle is enough

DHMO

15:33

Can any triangle be inscribed in a semicircle?

Balarka Sen

no

DHMO

for example?

Balarka Sen

literally anything. one of the interior angles have to be 180/2

SteamyRoot

Equilateral triangle?

DHMO

not really

If one angle is obtuse, then it is certainly possible

because the inscribed circle would have the center outside of the triangle

so the triangle would occupy less than a semicircle

Balarka Sen

15:35

oh i'm dumb yes

DHMO

@SteamyRoot the semicircle has its diameter touching the ground

make one tip of the equilateral triangle touch the center

never mind i'd just draw a diagram

Balarka Sen

for some reason i was assuming center of the circle is supposed to lie on a side but of course that's nonsense

@DHMO that is not inscribed. one of the vertices is not on the circle

SteamyRoot

^

DHMO

@BalarkaSen semicircle.

SteamyRoot

Semicircle is half a circle

DHMO

15:37

yes

Balarka Sen

semicircle doesn't contain that diameter

DHMO

what the hell

SteamyRoot

Not the boundary of half a disk

DHMO

that's new to me

the diagram on wiki includes the diameter

but the one on mathworld does not

Balarka Sen

read the definition, not the diagram

a semicircle is in particular an arc

DHMO

15:39

Euclid's definition includes it though

> A semicircle is the figure contained by the diameter and the circumference cut off by it. And the center of the semicircle is the same as that of the circle.

(The Elements: Book I: Definition 18)

Balarka Sen

that seems to indicate it's half a disk, not a one dimensional object we think of it as. nobody actually uses that defn

SteamyRoot

Euclid defines things differently

In particular, his circle is our disk

DHMO

that's interesting

SteamyRoot

and our circle is something he always describes as the "circumference of a circle"

DHMO

i learnt something new today

how do you describe the wrong semicircle then

I'm interested if every triangle can be inscribed in a wrong semicircle

SteamyRoot

15:41

Boundary of half a disk?

DHMO

let's just call it the wrong semicircle

@BalarkaSen thoughts?

SteamyRoot

Wouldn't that be kind of trivial?

Pichi Wuana

But, are all 90° angles that point to the diameter inscribed in the circle?

What I mean is that I have an ellipse

DHMO

@SteamyRoot so you mean yes?

why?

SteamyRoot

Look at the diagram you drew before

Pichi Wuana

15:44

SteamyRoot

Choose one angle to touch the diameter

make the opposite edge parallel to the diameter

and let the center of that edge be "right above" the center of the semicircle

Hmmm... actually, that might not work

You'll have to pick the right angle first.

Balarka Sen

@DHMO Well, equilateral triangle is a ctrexample

DHMO

@BalarkaSen it can be inscribed in the wrong semicircle

just as demonstrated above

Balarka Sen

@DHMO oh you want to do it with the wrong one. shrug.

SteamyRoot

Hmmm. My idea works if you pick the largest angle to be the one touching the diameter.

Pichi Wuana

15:51

In my ellipse D and E are the focal points. F is a point in the first quadrant and on the ellipse. DFE = 90°. Can I take DE as a diameter, $(0,0)$ as the center of a circle, and F as a point in the circle?

DHMO

@SteamyRoot nice. I tried it in geogebra

Pichi Wuana

And then FO = EO = DO

DHMO

@PichiWuana what is your question?

Pichi Wuana

If I can take that as a fact.

DHMO

you don't need the ellipse at all

Pichi Wuana

15:54

hmm

DHMO

@PichiWuana let O be mid point of BC

construct foot of perpendicular from O to AB as D, to AC as F

by definition, ODAF is a rectangle

OA is a diagonal of the rectangle, so ΔOAD = ΔAOF

i'm probably making everything complicated

by intercept theorem, CF=FA

by SAS, ΔAOF = ΔCOF

similarly, ΔOAD=ΔOBD

so ΔOBD = ΔOAD = ΔAOF = ΔCOF

so OB = OA = AO = CO

Secret

Finally
THE
****
IS
DONE
!

[Integral symmetries] Cauchy Riemann theorem like analogues in reals:

$$\frac{d^n}{dx^n}f=\lambda\int^{(m)}fdx$$

(where $f=\sin, \cos,e^x$ (possibly more?),$\lambda \in \mathbb{R}$)

For example

$$\int x^n \sin x e^x dx$$
Consider

$(uv)''=u''v+2u'v'+uv''$

Plug $u=\sin x$, $v=e^x$. Then

$(uv)''=-uv+2u'v+uv$

Note how the antisymmetry of $\sin x$ wrt the functional $()''$ results in cancellation. Hence

$(uv)''=2u'v$

Now it is easy to check that $\cos x$ is also antisymmetric wrt $()''$ thus we could have start with $u' =\sin x\implies u = -\cos x$. Hence

DHMO

@Secret what would fit in the asterisks... "fuck" doesn't fit there.

oh, "shit" fits there.

Secret

I have been working for 3 days on this ever since simpleart gave me that challenge. It should not took that long but I keep making careless mistakes after careless mistakes

DHMO

what is the challenge?

Secret

16:08

in Simply Beautiful Art's room, 2 days ago, by Simple Art

@Secret can you take this integral without IBP?$$\int x^2\sin(x)e^x\ dx$$

(Please let me type the nxt msg before responding)

For simpleart's challenge, I do managed to find a IBP free way to evaluate it, and then some careless mistake or something result in the integral to get wrong ans, but simpleart said it is fine as my ans agrees with the form of the ans of that integral

When doing his challenge, I got lazy (and at the same time I want to test my hypothesis on integrand symmetries) thus I generalise his integral to the one above

I then found (as you saw in that long wall of text), the n power version of that integral has a close form which is captured by the symmetry of the integrands

(The IBP free method to do simpleart's integral is the starred message of today)

Secret

NB whenever I got lazy, I went into generalisation mode to derive a universal formula for all types of problems similar to the problem I am working in. However because generalisation proofs are in general hard, paradoxically, it result in a lot more work than simply blowing through the problem directly

DHMO

@Secret I really like your passion.

Secret

Alternately, I have a weird metric on measuring effort. To me the more I need to repeat the exact same thing, the more effort it takes

Sophie

16:32

can someone prove $\int_{-1}^0 \frac{e^x+e^{1/x}-1}{x}dx=\gamma$?

DHMO

@Sophie trapezoidal rule?

Sophie

what?

DHMO

Riemann sum

In mathematics, a Riemann sum is an approximation that takes the form ∑ f ( x ) Δ x {\displaystyle \sum f(x)\Delta x} . It is named after German mathematician Bernhard Riemann. One very common application is approximating the area of functions or lines on a graph, but also the length of curves and other approximations. The sum is calculated by dividing the region up into shapes (rectangles, trapezoids, parabolas, or cubics) that together form a region that is similar to the region being measured, then ...

basically the definition of definite integral

Sophie

I know that $\int_{-1}^0 \frac{e^x-1}{x}dx=\sum_{h=1}^\infty \frac{(-1)^{h+1}}{h\cdot h!}$

@DHMO I know this, but how do you use it to prove the integral is exactly $\gamma$? I already know it is pretty close numerically, which is why I'm wondering

DHMO

@Sophie just an idea; I haven't tried it

Secret

16:38

Got a very simialr looking expression in this page

Exponential integral

In mathematics, the exponential integral Ei is a special function on the complex plane. It is defined as one particular definite integral of the ratio between an exponential function and its argument. == Definitions == For real non zero values of x, the exponential integral Ei(x) is defined as Ei ⁡ ( x ) = − ∫ − x ∞ e − t ...

In fact the $\int \frac{e^{x^{-1}}}{x}dx$ bit migh be related to the exponential integral

Sophie

this came up because I think $\int_0^x\frac{e^x-1}{x}dx\sim \gamma-\ln(-x)$ as $x\to -\infty$

Secret

Sophie

that's what I'm looking for, I think

Sophie

16:59

this guy says there's no perfect cuboid arxiv.org/abs/1506.02215

DHMO

@Sophie isn't this one of the million dollar questions?

Sophie

nope

DHMO

no?

right

Secret

So if there is no perfect cuboid, then it should not be possible to have a perfect hypercuboid since at least one of its (n-1)D faces will need to be a perfect cuboid

DHMO

@Secret there you are, generalizing again :p

Mike Miller

17:05

There's no particular reason to believe that paper.

Alessandro Codenotti

There are papers on the arvix claiming all sort of outrageous things from P=NP to the Goldbach conjecture...

Secret

Well, I am not even sure what's the significance of (non)existence of parallepippeds where the diagonals are all integers

DHMO

cuboids @Secret not parallelpipeds

Secret

that paper define 'cuboids' as some kind of parallelpipped

Goldbach's conjecture

Goldbach's conjecture is one of the oldest and best-known unsolved problems in number theory and all of mathematics. It states: Every even integer greater than 2 can be expressed as the sum of two primes. The conjecture has been shown to hold up through 4 × 1018, but remains unproven despite considerable effort. == Goldbach numberEdit == A Goldbach number is a positive integer that can be expressed as the sum of two odd primes. Since four is the only even number greater than two that requires the even prime 2 in order to be written as the sum of two primes, another form of the statement ...

What happens if we can show there exists integers that are not sum of two primes?

Astyx

The Goldbach Conjecture is disproved ?

Sophie

17:15

@MikeMiller why not?

PVAL-inactive

That's a 45 page paper with essentially no explanation, no sketch of the proof just random likely meaningless equations

Lucas Henrique

Hello there

Secret

No I mean: Suppose there exists integers that are not sum of two primes, what does that mean?

Alessandro Codenotti

17:33

Probably nothing. The interesting part will be the techniques used to prove that there is (or there isn't) such an integer

Secret

I see

Lucas Henrique

Are any of you guys acquainted with graphs? (graph theory)

Sophie

does $\int_\pi \frac{1}{t}dt=\ln(x)$ for a real positive number $x$ for any contour $\pi$ that goes from 1 to $x$ in the complex plane?

DHMO

@Sophie good luck passing through the origin

if $\pi$ doesn't pass through the origin, then yes.

the origin is the only pole of $f(z) = \dfrac 1z$.

Sophie

@LucasHenrique this is like asking someone if they're familiar with number theory. Maybe they'll say yes and then you'll ask something bizarre about inter-universal Teichmuller theory. Maybe they'll say no and you'll ask an introductory question they did know how to answer. So if you state the question you're more likely to get an answer

arctic tern

17:45

@Sophie no

it will be ln(x) + 2pi i times the number of times the contour goes around 0 counterclockwise

Lucas Henrique

Ok then :p

My problem is the following: I have, for each vertex of a graph, the set of vertex it's connected to

I want to build a graph that is closed given this information

How can I do this?

s.harp

What do you mean with "closed"?

Lucas Henrique

Eh... I want to build "a circle"

(I have neither vocabulary nor theorical basis on graphs, the book just solves problem using graphs without an introduction)

A cyclic graph

s.harp

If your graph is not directed you and for $v$ a vertex $N(v)$ is the set of vertexes connected to it you get the set of edges to be $\{ v\to w \mid w\in N(v) \}$

Sophie

if you have the set of vertices any vertex is connected to then your graph is already determined

s.harp

17:50

Knowing what vertices are connected to what other vertices completely describes the graph

Sophie

that's how you define a graph. The dots and lines interpretation is "just" a representation

Lucas Henrique

I don't know how can I represent my graph tho

I need to prove that, given the set of vertices, it's possible to create a cyclic graph

(I know it's a cyclic graph)

It's given by the table (each number is corresponding to a different graph)

Secret

[Random Rambles]
Suppose the Goldbach conjecture is true. Then for all even integers $2n \in \mathbb{Z}^+$, there exists $m,r \in \mathbb{Z}^+$ such that it can be wrote as $2n=(2m-1)+(2r-1)=2(m+r-1)$, where $2m-1$,$2r-1$ are primes.

Rewriting the equation, we get

$n=m+r-1$

$m-r=-(n+1)------(*)$

which is an equation of a 3D linear subspace in $\mathbb{R}^4$

By taking cross sections $n \in \mathbb{Z}^+$, the problem becomes the above equation always have a solution in terms of integers $m,r$ such that they corresponding to the aforementioned primes

s.harp

If you just have vertices take the graph so that all vertices are connected to each other

Lucas Henrique

$1: 6, 7, 9\\
2: 3, 9, 10\\
3: 2, 9, 11\\
4: 10, 12\\
5: 7, 11\\
6: 1, 8, 12\\
7: 1, 5, 12\\
8: 2, 6\\
9: 1, 3\\
10: 2, 4, 11\\
11: 3, 5, 10\\
12: 4, 6, 7$

I want to prove that it's possible to find a path that it's possible to walk at every vertex exactly one time and end in the first vertex

Sophie

17:59

draw the graph on a piece of paper and it shouldn't be hard

Astyx

Hi @Ted

Lucas Henrique

@Sophie I have been trying... but it is hard :|

Ted Shifrin

Hi @Astyx

Astyx

Comment vas-tu ?

Secret

Actually is $gcd(1,-1)$ well defined...?

Astyx

18:02

It's 1 @Secret

Sophie

you could write a program to do it. Shouldn't be hard, but maybe it is

Ted Shifrin

Ça va, merci, et toi?

Balarka Sen

Hi @TedShifrin

Ted Shifrin

Hi @Balarka

Astyx

Ça va ça va, plus que semaines mois avant les concours donc la pression monte

Secret

18:03

Hmm...

> This Diophantine equation has a solution (where x and y are integers) if and only if c is a multiple of the greatest common divisor of a and b. Moreover, if (x, y) is a solution, then the other solutions have the form (x + kv, y − ku), where k is an arbitrary integer, and u and v are the quotients of a and b (respectively) by the greatest common divisor of a and b.

Ted Shifrin

Il faut toujours de la pression :P

@Balarka: What hast thou to report?

Danu

Hi Ted

Ted Shifrin

Hi @Danu ... I'm back after 7 1/2 hours of no electricity last night :(

Danu

Oh yikes

Astyx

@Secret To me $\gcd(a,b)$ was defined as the only positive integer $g$ such that $g\Bbb Z = a\Bbb Z + b\Bbb Z$

Danu

18:04

A good omen, I'm sure :)

In any case, I'm kind of in despair with a calculation that we talked about last week :(

Ted Shifrin

When I worked on my thesis, I spent months at a time stuck/confused!! :)

Astyx

@Ted Oui c'est sûr, enfin ce qui m'ennuie un peu c'est d'être "obligé" de faire de la physique, du français et de l'anglais pour les concours alors que je préférerais que des maths. Mais bon, c'est la vie :)

Secret

$m-r=-(n+1)$ is a linear Diophantine equation of the form $ax+by=c$ where $a=1$,$b=-1$ and $c=-(n-1)$. Since $gcd(a,b)=1$, $c$ is trivially a multiple of the gcd for all $n \in \mathbb{Z}^+$. Therefore this equation always had a solution (as expected for linear subspaces as they span the whole codomain).

The question however is defining a lattice of primes to restrict the values of $m, r$

Danu

@TedShifrin I'm off by a sign at least in one place---but the sign problem arises from stuff I don't understand so I've been unable to correct it.

Ted Shifrin

A mon avis, il faut apprendre plus que les maths pour avoir une bonne éducation :)

Danu

18:07

LA PHYSIQUE

Ted Shifrin

@Danu: This is why topologists like $\Bbb Z_2$ classes :P

Danu

@TedShifrin I always felt $\Bbb Z_2$ was a cop-out.

Ted Shifrin

Well, sometimes it's all you got :P

Danu

I guess I'll have to go back to Kotschick and ask him to explain :(

Mahmoud

Hi.

Astyx

18:09

Le fait est que tous les cours de maths que je reçois sont du même prof pour un programme assez conséquent (concours obligent), et donc on ne fais que "survoler" des notions de maths (j’exagère un peu)

Enfin bon, je n'ai cas arrêter de me plaindre et réussir les concours :p

HI @Mahmoud

Ted Shifrin

Evidemment, tu vas mieux mais il te faut te plaindre quand même :P

@Astyx: Pity the students who were stuck with me for 4 or 5 classes in their college careers :D

Astyx

N'est pas français qui veut ! :p

Ted Shifrin

Hi @Mahmoud

Sophie

reading math in german is far easier than I expected

Ted Shifrin

Once you know the key words in the subject you're reading about, @Sophie.

Buongiorno @Alessandro

Mahmoud

18:14

Il faut se plaindre pour que tu n'oublies jamais que les concours no reflectent pas le noyau des Mathématiques. @Astyx

Socrates

Where can I read about inducing an order?

Astyx

qu'à* (see three messages above)

Can't believe I made that mistake

Ted Shifrin

Hell if I can count all your messages.

I guess Balarka is ignoring me ... I shouldn't complain.

Danu

@Astyx Good! I saw that sentence and did a triple-take, then decided my knowledge of French was insufficient.

Mahmoud

$$\lim_{x\to +\infty} f(x)=L \iff (\forall \epsilon)(\exists \delta)(\forall x\in D_f) : x\gt \delta \implies |f(x)-L|\lt \epsilon$$ @TedShifrin I feel (more of think*) that the part of $x\gt \delta$ isn't enough to ensure that $x$ gets unbounded.

Danu

18:20

But I guess I wasn't wrong :D

Astyx

My English has become better than my French it seems :p @Danu

Ted Shifrin

I would use a letter like $M$ rather than $\delta$, but it's right, @Mahmoud. If you say $x>1000$, then certainly $x$ must be allowed to go off to infinity.

Mahmoud

@TedShifrin But the quantifier is $\exists$

Astyx

How does one show $GL_n^+(\Bbb R)$ is connected ? Is using transvections and dilatation matrices with a positive coefficient a good way ? What other proofs are there ?

Ted Shifrin

It's like saying $|x-a|<0.1$ doesn't force $|x-a|$ to be tiny, tiny. But as you shrink $\epsilon$, you will presumably get $\delta$ smaller (or $M$ larger in the infinity case).

The easiest way is the $QR$ decomposition, @Astyx. You can write every invertible matrix (uniquely) as the product of an orthogonal matrix and an upper-triangular matrix with positive diagonal entries.

Mahmoud

18:23

@TedShifrin Okay thanks, I'll see some concrete cases. $:)$

Ted Shifrin

@Mahmoud: It's the exact same logic as for the usual limit definition.

Astyx

Do orthogonal matrices form a connected set ?

Ted Shifrin

Ones of positive determinant do.

Danu

Naw

Ted Shifrin

The easiest way to prove this is to give a normal form for orthogonal matrices. The easiest way I know to do that uses the spectral theorem, but you can do an induction argument.

Danu

18:26

To see it, look at the (continuous) function $\det$

@TedShifrin Why not ^

Astyx

Yes, I meant positive determinant @Danu

Ted Shifrin

I said positive determinant 5 lines up, @Danu.

Astyx

(which I should have explicited)

Danu

Oh, sorry.

Ted Shifrin

Glad to see @Danu is obsessed with minus signs :P

Astyx

18:27

It's my fault really

Danu

@TedShifrin Hmm... is it easy to prove that $GL^+$ is connected? Because then Gram-Schmidt is very elegant

Ted Shifrin

I'm suggesting this approach to prove precisely that, @Danu. The $QR$ decomposition is based on Gram-Schmidt.

Semiclassical

Hi chat

Astyx

Hi @Semiclassical

Danu

@TedShifrin I forgot what all those decompositions mean :P

Socrates

18:28

what is inducing an order?

Ted Shifrin

But I explained it, dammit.

Hi @Semiclassic.

Semiclassical

Will probably have questions for you later, @ted. (On mobile right now)

Danu

@TedShifrin I didn't read back :P

s.harp

A topological space is of finite type if only a finite number of homology groups are non-zero?

(Thats a question about what the definition is)

Alessandro Codenotti

Hi @Ted

Astyx

18:33

Is the spectral theorem : "All symmetric real matrices are similar to a diagonal matrix." ?

Socrates

@BalarkaSen imgflip.com/i/1i385c

Ted Shifrin

@Astyx: You need the generalized version for normal operators.

Astyx

Right, haven't done that yet

Alessandro Codenotti

It also tells you that there is an orthonormal diagonalizing basis made of eigenvectors

Ted Shifrin

You can do an induction argument. Look for invariant $2$-dimensional subspaces on which it is a rotation.

Astyx

18:37

Yes, it's quite obvious that $SO_2(\Bbb R)$ is connected

Since we have a parametrization

Ted Shifrin

I'm claiming you can find a basis in which it looks like a bunch of such $2\times 2$ blocks, and then $\pm 1$ in the remaining diagonal slots.

Astyx

For the general case $SO_n(\Bbb R)$ ?

Ted Shifrin

Yes.

Astyx

Okay, I'll think about it after supper

Bye !

Ted Shifrin

Bubye.

Balarka Sen

18:51

@TedShifrin Nothing much to report. I think I do see how to prove G-B using a Poincare-Hopf argument though.

@Socrates Nice!

@s.harp Maybe it means it's homotopy equivalent to a finite CW-complex?

nullgeppetto

Hi all! I have a couple of questions regarding a concentration inequality about a function of normal variables. Could anyone help? Thanks?

So, I leave it here; if anyone could take a look, that would be great!

Let's suppose that $X=(x_1, \ldots, x_n)$ is a vector of i.i.d. (standard) Gaussian variables and $f\colon\Bbb{R}^n\to\Bbb{R}$ is L-Lipschitz wrt to the Eucledian norm. Then, the following concentration inequality holds for all $t>0$: $$P(\lvert f(X)-\Bbb{E}[f(X)]\rvert\geq t)\leq 2\exp(-\frac{t^2}{2L})$$.

My basic question is the following: Suppose that I have a formula for computing $\Bbb{E}[X]$, without the need of sampling from the Gaussian. Now, I would like to know, how many samples I should draw from the Gaussian, so that the estimation I obtain for the (sample) mean, is approximately the same as that I get from my formula. As you may imagine, I'm interested in a relation between the sampling size (say $N$) and the dimensionality $n$.

I'm not sure, though, if the concentration inequality can help.

Thanks for taking the time reading my question!

Typo: I meant a formula from computing $\Bbb{E}[f(X)]$...

Perturbative

19:17

Hey everyone!

Lucas Henrique

@Perturbative Hello there

Perturbative

In $3(b)$ above, since $X$ and $X'$ are just 'symbolic', does the result that the topology on $X'$ is finer than on $X$ essentially assert that the topologies induced by the metric $d$ on $X'$ and $X$ are equivalent?

Astyx

@Ted My initial idea was that, as a corrolary of the Gaussian elimination, any matrix of determinant 1 can be written as a product of $T_{i,j}(\lambda) = I_n + \lambda E_{i,j}$ for $i\ne j$ and $\lambda \in \Bbb R$.
So to show $GL_n^+(\Bbb R)$ is connected, you only need to show that for any invertible matrix of positive determinant $P$ you can go continuously from $I_n$ to $P$ in $GL_n^+(\Bbb R)$.
So write $P$ as $$P = \det(P)T_{i_1,j_1}(\lambda_1) \dots T_{i_n,j_n}(\lambda_n)$$
And consider $$P(t) =((1-t) + t\det(P))T_{i_1,j_1}(t \lambda_1) \dots T_{i_n,j_n}(t \lambda_n)$$. We have $P(0) …

Mahmoud

19:38

Hi.

This should have been simple to understand, but I'm still struggling.

Mary Star

Let $S_n=H_0\triangleright H_1 \triangleright \ldots \triangleright H_r=\langle 1\rangle$ and $H_{i-1}/H_i$ be abelian.

Why can the above not hold when each $H_i$ contains all the cycles of length $3$ ?

Daminark

Does that make more sense?

And @Perturbative that's not exactly what it's asking. $X'$ is denoting a space, so that means we've already specified some topology on it. What the question is saying is that any set that was open under the metric space topology generated by $d$ is also open under the pre-determined topology of $X'$

The way you said it was a bit off, because a metric isn't defined on a topology, it's defined on a set, and can generate a topology, so now we're asking to compare two topologies on the same set.

Mahmoud

19:55

@Daminark Why did it have to be $0\lt x\lt 2a$ ?

Daminark

Well, $|x| < y$ means that $-y < x < y$. So applying it to this particular scenario, $-a < x-a < a$. Well, if you add $a$ to the expression, you get $0 < x < 2a$

Mahmoud

I know right, but I don't see why did it make the implication hold.

In which $a$ plays the role of $\delta$

@Daminark

Daminark

The point is that it makes $x > 0$. By the definition of the function, this means that $f(x) = 1$. Well, $f(a) = 1$ as well, so that $|f(x) - f(a)| = 0$, which is less than any $\epsilon > 0$ you could've chosen.

Mahmoud

@Daminark Thanks.

Liad

let $ f : I \to R \ ^ n$ be a continuously differentiable function. $ I \subset R$ open interval.

Define $g: I \ ^ 2 \to R \ ^ n $ by $ g(x,y) = Df(x)$ if $ x = y$ and $ g(x,y ) = \dfrac{f(x)-f(y)}{x-y}$ else.

I need to prove that $ g$ is continuously differentiable in $I \ ^ 2 {- \{ (x,x) : x \in I \}} $ and continuous in $I \ ^ 2 $. im not sure how to do that..someone see a way?

Liad

20:42

never mind. solved it :-)

Display name

Latex is not compiling in this chat

I'm still seeing the raw code

Semiclassical

use the latex in chat link in the room desc

Astyx

See top right

Display name

ok it's working now

Astyx

Good

Socrates

21:08

so appearantly one can order any set with "negative" elements (not neccessarily negative in the usual terms)

Hawk

is any1 here?

Socrates

no

Tobias Kildetoft

@Socrates I am not sure what you mean by negative there

Socrates

additive inverses

Tobias Kildetoft

@Socrates What does that have to do with ordering?

Socrates

21:10

mmh, maybe I misunderstood

Salehen Rahman

I'm curious about this: given some polynomial $p$ of degree $m$, does $p^k$ yield a polynomial of degree $m \times k$?

Hawk

If $R/I$ is a quotient ring with ideal I in R, then can I define the set ${a + I: a \not\in I}$ to be an ideal of $R/I$?

Tobias Kildetoft

@SalehenRahman Yes

@Hawk That will not be an ideal

Salehen Rahman

@TobiasKildetoft Thanks! Where can I read more about this?

Hawk

But why not?

Tobias Kildetoft

21:12

@Hawk Try any example to see that

@SalehenRahman This follows from the fact that the degrees get added when you multiply polynomials

Socrates

@TobiasKildetoft math.stackexchange.com/a/2107749/346682 this only works with additive inverses

Salehen Rahman

@Tobias, right, thanks!

Tobias Kildetoft

@Socrates So what you really wanted to say was that knowing which elements are positive leads to an ordering when you have additive inverses (and require that the ordering is compatible with addition)

Socrates

I frankly don't know what I wanted to say. I don't understand this. Is $\mathbb{N}$ not orderable?

Tobias Kildetoft

of course it is.

Hawk

21:27

Okay I i just tried it on $Z/2Z$ and ${1 + 2Z}$. First $1 \not\in 2Z$ and second we see that $2Z(1 + 2Z) = 2Z$ and $(1 + 2Z)(1 + 2Z) = 1 + 2Z$. Meaning $Z/2Z{1 + 2Z} = {2Z, 1 + 2Z} \not\in {1 + 2Z}$. But I tried applying this to this proof I am reading on $R/M{a + M}$ and they claim it is an ideal. Am I missing something? imgur.com/a/1DmWh. I know they are assuming $a + M$ has no inverse.

For some context, I am writing the proof without using R/M is a field and using the alternate condition that R/M contains only 2 trivial ideals and I am not using any correspondence theorems

Tobias Kildetoft

@Hawk The thing they claim is an ideal is the one generated by $a + I$ in the quotient, not the set of all elements of that form

(not sure why one would go through this proof before the correspondence though, as that makes it much more conceptual)

Display name

@hawk If you observe that 2Z is the set of "even integers" and 2Z+1 is the set of "odd integers" then it's easy to see how the equations hold

towc

any LC fans here?

also, anyone who gets paid for being a mathematician?

not teachers

Tobias Kildetoft

21:43

What's LC

and I used to get paid (and will again in about half a year)

towc

kinda curious what level of skill does one need to be competitive and earn a decent amount, by actually doing things one loved

@TobiasKildetoft if you were an LC fan, you'd know ;)

lambda calculus

I just bloody love that world mate

I wish I could float in lambdas and be happy

Tobias Kildetoft

Not sure how much the salary really depends on skill, as most universities have fairly fixed salaries

towc

again, not teachers

and if you mean uni researcher, then maybe

Tobias Kildetoft

Right, researcher

towc

but uni research is not what I'm looking for

Tobias Kildetoft

21:45

nobody else wants to pay me for my research

towc

I'm a web developer by day, and an LC dreamer by night

well, not for personal research

which is why I was wondering what other options are out there

Astyx

It's simple, you just have to solve one of the millenium problems

towc

yeah, right -_-

wait until trump says he's the world's greatest mathematician

Balarka Sen

rehi @Ted

Ted Shifrin

rehi @Balarka

Socrates

21:49

@TedShifrin do you mean with trichonomy "either a<b, b>a or a=b"?

Ted Shifrin

Yes, @Socrates, with exclusive or.

Oops, no, $a<b$, $b<a$, or $a=b$.

Socrates

and why does this fail with 0 being negative and positive? Oh, yes, a typo!

Astyx

@Ted I guess you didn't read my proof ? :)

Ted Shifrin

exclusive or, @Socrates

Precisely one of the three occurs.

Nope, Astyx. What proof?

Astyx

This one

Ted Shifrin

21:52

Oh. Wait a minute.

Socrates

ah, if we say 0 is positive and negative, that implies 0<0 and 0=0, or how you mean?

Astyx

You don't have to, don't waste time on me !

Ted Shifrin

So I'm not sure I know that your first step is true, @Astyx.

Astyx

Using only transvections one can always diagonalize any matrix and have only 1's in the diagonal except $a_{n,n} = \det(A)$. Then using transvections again one can triangularize any matrix.

Ted Shifrin

@Socrates: I guess I want to say precisely one of the following occurs: $x$ is positive, $x$ is negative, or $x=0$. ... If your definition of "positive" is $x\ge 0$, then that violates precisely one.

Socrates

21:55

oh, I didn't thought this way

Ted Shifrin

I guess I've never thought through this, @Astyx. So how do you do $\begin{bmatrix} 0 &-1 \\ 1 & 0\end{bmatrix}$ and $\begin{bmatrix} 2 & 0 \\ 0 & 1/2\end{bmatrix}$?

Hawk

@TobiasKildetoft somewhat unrelated, but in the proof I linked. Because $a \in R - M$ (R without M), then $r{a}$ is a principal ideal?

Astyx

For the first matrix : Add the bottom line to the first. Remove the first from the bottom one. Add the bottom one to the first.

Tobias Kildetoft

@Hawk $Ra$ is always a principal ideal, by definition

Alessandro Codenotti

@Tobias I'm a bit confused by the correspondence theorem (I just googled it after you mentioned it), so I see how the existence of a bijective map from subgroups of $G/N$ to subgroups of $G$ containing $N$ follows from the third isomorphism theorem, I'm not sure what it's meant with "those $2$ sets have the same structure"

Display name

21:59

@Socrates You need to assume that for positive A and negative B, A>0 B<0 and A>B. For any positive A you can find a smaller positive A'=A/2 and for any negative B you can find a larger B'=B/2. Then since 0/2=0 you get 0>0 and 0<0 giving contradictions in both cases

Tobias Kildetoft

@AlessandroCodenotti That probably refers to their structure as a poset

also, the correspondence preserves normal subgroups

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